Решите неравенство f`(x)>0, если f(x)=cos4x+2x

Question 1

Question 2

0;\ \ x-?;\\ f'(x)=\left(\cos4x\right)'+\left(2x\right)'=-4\sin4x+2;\\ f'(x)>0;\\ -4\sin4x+2>0;\\ \sin4x=t;==>-1\leq t\leq1==>\\ t\in[-1;1];\\ -4t+2>0;\\ 4t-2<0;\\ 4t<2;\\ t<\frac12;\\ -1\leq t<\frac12;\\ \sin4x<\frac12;\\ " alt="f(x)=\cos4x+2x;\\ f'(x)>0;\ \ x-?;\\ f'(x)=\left(\cos4x\right)'+\left(2x\right)'=-4\sin4x+2;\\ f'(x)>0;\\ -4\sin4x+2>0;\\ \sin4x=t;==>-1\leq t\leq1==>\\ t\in[-1;1];\\ -4t+2>0;\\ 4t-2<0;\\ 4t<2;\\ t<\frac12;\\ -1\leq t<\frac12;\\ \sin4x<\frac12;\\ " align="absmiddle" class="latex-formula">

$\frac{5\pi}{6}+2\pi n<4x<\frac{13\pi}{6}+2\pi n,\ n\in Z;\\ ili \\ .\ \ \ \ \ \ \ -\frac{7\pi}{6}+2\pi k<4x<\frac\pi6+2\pi k;\ k\in Z\\$

$\frac{5\pi}{24}+\frac{\pi n}{2}<x<\frac{13\pi}{24}+\frac{\pi n}{2},\ n\in Z;$
ili
$-\frac{7\pi}{24}+\frac{\pi k}{2}<x<\frac{\pi }{24}+\frac{\pi k}{ 2},\ \ k\in Z$

Question 3

а что после син4х<1/2 ?

мохинсан_zn · Answer 1 · 2018-04-01T07:03:56+0000

0;\ \ x-?;\\ f'(x)=\left(\cos4x\right)'+\left(2x\right)'=-4\sin4x+2;\\ f'(x)>0;\\ -4\sin4x+2>0;\\ \sin4x=t;==>-1\leq t\leq1==>\\ t\in[-1;1];\\ -4t+2>0;\\ 4t-2<0;\\ 4t<2;\\ t<\frac12;\\ -1\leq t<\frac12;\\ \sin4x<\frac12;\\ " alt="f(x)=\cos4x+2x;\\ f'(x)>0;\ \ x-?;\\ f'(x)=\left(\cos4x\right)'+\left(2x\right)'=-4\sin4x+2;\\ f'(x)>0;\\ -4\sin4x+2>0;\\ \sin4x=t;==>-1\leq t\leq1==>\\ t\in[-1;1];\\ -4t+2>0;\\ 4t-2<0;\\ 4t<2;\\ t<\frac12;\\ -1\leq t<\frac12;\\ \sin4x<\frac12;\\ " align="absmiddle" class="latex-formula">

$\frac{5\pi}{6}+2\pi n<4x<\frac{13\pi}{6}+2\pi n,\ n\in Z;\\ ili \\ .\ \ \ \ \ \ \ -\frac{7\pi}{6}+2\pi k<4x<\frac\pi6+2\pi k;\ k\in Z\\$

$\frac{5\pi}{24}+\frac{\pi n}{2}<x<\frac{13\pi}{24}+\frac{\pi n}{2},\ n\in Z;$
ili
$-\frac{7\pi}{24}+\frac{\pi k}{2}<x<\frac{\pi }{24}+\frac{\pi k}{ 2},\ \ k\in Z$